Catch David on the Numberphile podcast: trvid.com/video/video-9y1BGvnTyQA.html

Professor David Eisenbud on the infamous Collatz Conjecture, a simple problem that mathematicians may not be "ready" to crack.

More links & stuff in full description below ↓↓↓

Extra footage from this interview: trvid.com/video/video-O2_h3z1YgEU.html

Prof Eisenbud's 17-gon: trvid.com/video/video-87uo2TPrsl8.html

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7 Ağu 2016

Çalma listem

Daha sonra izle

Numberphile 3 yıl önce

Catch David on the Numberphile podcast: trvid.com/video/video-9y1BGvnTyQA.html

jeremiah mint 4 aylar önce

If you answer the problem in sets it works, 3× 1 set=9 and plus + 1 = 11 not 10

jeremiah mint 4 aylar önce

If you answer the problem in sets it works, 3× 1 set=9 and plus + 1 = 11 not 10.

jeremiah mint 4 aylar önce

@Tejas Pathak yep

jose rivera 5 aylar önce

2+1=4 If you draw an addition symbol ➕ then draw the top of a number 1 and connect it to the northern point then draw the foot of the number 1 at the southern point, then draw the hook and 45°sharp angle of the number 2 and attach it to the western point of the addition sign then you have what makes 2+1=4

Kris Barber Yıl önce

Well if it leads to an Infinite number why can't we just say n=... then it goes all the way up and all the way back down?

Scott Anderson 4 yıl önce

“Seven seems to be an odd number.” That’s why he’s a maths professor.

Widow 3 aylar önce

@Will K I'm no mathematician, but 16 seems "really even" to me because it leads to a even number 3 times in a row.

Canh Nguyen 4 aylar önce

@BurgerAdam🇺🇦 are u for real:))

Film Feline Yıl önce

Seven is the number of completeness and perfection (both physical and spiritual). It derives much of its meaning from being tied directly to God's creation of all things. The number 7 is also important in Hinduism, Islam and Judaism

The Playing Dutchman Yıl önce

@powerplay.55 Isn't prehaps the opposite of posthaps?

AtheismOP Yıl önce

@powerplay.55 Turn on ur subtitles & u will no more hear "prehaps".....

Pietro Gagliardi 5 yıl önce

My first introduction to the problem was one of the example in former Bell Labs Unix co-pioneer Jon Bentley's book "Programming Pearls", which is a collection of columns on the theory and practice of software design he wrote for seminal comp sci journal Communications of the ACM over the years, expanded and with exercises. One of the exercises in one of the early chapters was this conjecture (in the Second Edition it's Column 4 Question 5). In the back of the book is a collection of hints. Here is the hint for this question. It has stuck with me ever since I first read it: "If you solve this problem, run to the nearest mathematics department and ask for a Ph.D."

Red X 2 aylar önce

I have solved this problem. I can prove it.

Erik 6 aylar önce

@Tommy Encrapera it was with the 3n+1 that this interesting result was found (which wasn’t solvable) Btw if u take any even number for the coefficient (like 4n+1) everything explodes, since anything multiplied by an even number added by one is odd, and so forth

UniverseNerd 8 aylar önce

@Tommy Encrapera yes exactly

Tommy Encrapera 9 aylar önce

What is the formula trying to decipher? Like why did they choose those numbers 3x+1 and not 4x+1 or 3x-1 and so on , what is the end game ? Is it the Only formula that always falls back to 1 and they dont know why? I'm trying to figure out what it would prove if there is a number that goes onto infinity or completely forms a separate loop , then what would that determine? Sorry for not understanding and seeing what I'm missing .

Alexander Townsend Yıl önce

@elrak0 2 n could be negative, but then it wouldn't be the Collatz Conjecture. A conjecture is an educated guess in math. The collatz conjecture was an educated guess by the mathematician Lothar Collatz. The conjecture is something like, "I think that every positive whole number n goes to one under this process". While these are probably not his exact words you get the idea. He said nothing about n being negative. You could ask about negative numbers, but that is its own separate, but still interesting problem. I hope that clears things up.

Rebecca Gaskins 5 yıl önce

I'm reminded of the wikipedia phenomenon where, if you click the first link in almost any article that isn't in parenthesis or italics, and keep doing that long enough, you will eventually end up at philosophy.

Steverson 9 gün önce

Social science is a loop

AerialExplorer 21 gün önce

How interesting. I tried it: Landlord to philosophy in 26 clicks.

Noname Aylar önce

@Mysteryskatin after philosophy, it will stuck in a loop. Philosophy -> Existence -> Reality -> Object of the mind -> Object -> Philosophy

Plissken 2 aylar önce

I tried George Clooney on my first go of this. Spectacular counterexample.

Mysteryskatin 4 aylar önce

@DancingDoughnut It's only true if you stop at philosophy. The topic is arbitrary, if you click randomly eventually you will end up at whichever topic you desire.

Jalepeno PVP 3 yıl önce

This guy has legit the most calming and comforting voice I’ve ever heard

Artur B Yıl önce

He reminds me of my university professor so it actually makes me nervous as it reminds me about the work I have to do damn

Keystone Science 2 yıl önce

If a number is to not reach one, a constraint is that it must be congruent to 63 mod 96.

Gameson 7 aylar önce

I cant find anything saying that this is true

Peter Samuelson Yıl önce

@Orion Hunter Congruent means equal, but under some more specific operation / transformation. In this case, they're saying that if any number violates the Collatz conjecture, that number mod 96 will be equal to 63. By "mod 96", they mean if you divide the number by 96, the _remainder_ will be 63. One could also write it as 63 + 96n where n is a nonnegative integer.

Yash05 Yıl önce

@Levent K******** man i've seen you in every comment section ever

Levent K******** Yıl önce

@Penta Idek I existed 11 months ago

Penta Yıl önce

@Levent K******** It's the chess comment section guy

GWS 4 yıl önce

Extremely interesting! Just going over it I can already kinda see why this is a problem. You would have to find a set of numbers either where 3n+1/2 appears more than n/2, or you'd have to find a set of numbers that creates its own tree, which would probably be an extraordinary large number considering that, as numbers increase, it's less likely that you'll get a cyclical sequence.

Alan Tyte 4 yıl önce

A cyclical sequence is inevitable as there are only six rules for finding the pivot of the 1-chain which effectively boil down to 3. The length of the cycle triples at every step up. The numbers do get huge but are consistent. Easier to talk directly.

esotericVideos 5 yıl önce

The first part of every numberphile video gets me excited to try to solve a problem, the second part convinces me that there's no point in trying because tons of people have already tried. :/

Simon Reinsperger 5 yıl önce

"Wow... 16, very even number!"

Harmonica Bruce 3 aylar önce

Even numbers can be divided by 2. 16 can be divided by 2 three times, so it's very even.

Edward Schmidt 11 aylar önce

I loved that!! I know there are not degrees of evenness. A number is either even or its not. Yet I felt that immediately when he said it. I smiled all the way from childhood at that comment. Some numbers have just always "felt" more even or more right to me. Not just the 2 to the X numbers. Also numbers like 12 or 24. So many like the low prime numbers. People love 3s and 7s. Not me. I love the numbers that can break apart clean so many ways. The numbers that make me go "Wow...that's a very even number!".

Sz. 11 aylar önce

@FCIG 0504 Hehh, amazing! :)

Linus Berglund Yıl önce

Any programmer will tell you that some numbers are more even than others. 16 in indeed very even.

Jim Cullen 6 yıl önce

Man I chose 4 initially. Worst possible number…

TTR SINGAM 9 aylar önce

I tried it with all numbers between 1 to 10 and......

Sweet Durt Yıl önce

1

Hiba Abdul Salam Yıl önce

@Roy Lavecchia You can consider positive integers only and also numbers ≥ 2

WerewolfLord Yıl önce

"Choose a number between 1 and 10." Ok, I'll pick e.

Yadu K Yıl önce

Here's a tip: don't choose a power of 2👍

Luke Blankenberg 3 yıl önce

It seems to me (although this has probably been pointed out many times) that just because (3n+1)/2 > n we do not have to show that there are more n/2's in order to reach 1. What we have to do is show that it hits some power of 2 (as then it will immediately go to 1)(the last 1/2 should be fine as hitting a power of 2 or the next power of 2 is equivalent for the purposes of going to 1). The issue then is whether this is the limiting case (all cases with more n/2's in them before a power of 2, so that there aren't equal amounts of them, will all go to a power of 2 if this one does/they will eventually goes on a track that goes to this alternating between even and odd sequence that we might be able to show goes to a power of 2, or if some do not necessarily do so and so go to infinity). However, I have a nasty feeling that actually showing this is pretty hard. But I thought I might as well throw this out there.

J C 2 yıl önce

My friend in high school told me about this nearly 20 years ago, so cool to see a video about it!

Hiba Abdul Salam Yıl önce

@Mega User The Conjecture is defined for Positive Integers ≥ 2

asia mies Yıl önce

@Mega User 0 not natural

John Smith Yıl önce

It's amazing how many teachers shouldn't be teachers , after just paying a few minutes of attention while I was eating I just posted the solution in the comments section.

Mega User Yıl önce

I notice that if you start with 0, you don't get to 1, but you keep looping back to 0 over and over again, since 0 is even. Also, at the end when they ask "Why not 3*n - 1?", I thought "Hey, if we do the 3*n + 1 problem with negative numbers, then if we let n = -m, where m is a positive odd number, then we get 3n + 1 = 3*(-m) + 1 = -3*m - (-1) = - (3*m - 1), and if m is even, then we get 2*n = 2*(-m) = -(2*m), so the negative cases reduce to the 3*n - 1 problem for positive n (if we take the absolute value)!"

Mia V Yıl önce

When he said any fourth grader could understand it, I was like, that's probably an exaggeration. Then he explained it and sure enough I remembered reading about it in a children's math book called "Math For Smarty Pants" in elementary school.

Matthew Doucette Yıl önce

Ha, awesome.

Albeit_Jordan 3 yıl önce

I think the key to this conjecture takes the form of yet another such - if we find the probability of *n* either being odd or even in the simple expression *x∙y=n* (where x and y are odd and even whole numbers) then I reckon it'd just be a game of permutations from there.

Faisal swaty 4 aylar önce

I am working on Prime numbers and I am amazed to see your expression they are extremely useful in my work Thank you man

Léo Bitencourt 5 yıl önce

I think it has to do with an odd number multiplied by 3 adding 1 always being an even number, but an even number divided by two could give ya an odd number or another even number, so it kinda eventually forces it into powers of 2 and we all know it's downhill from there

K B 7 gün önce

Yeah that's the intuition behind the "conjecture" but we can't prove that eventually it will reach powers of two.

Bengal Basi Yıl önce

The " collatz behavior " is first shown when converting an odd p into even using p+1 and repeating collatz process infinitely. This collatz behavior originates from 1(p)+1. This exact behavior is observed using 3p+1. But loses using 5p+1.

Alexander Korshkov 3 yıl önce

I used to think that there are only two possibilities: if you start with some number you either eventually get to 1 or the sequence would fluctuate but eventually grow to infinity. But there is one more possibility: the sequence might get to some other cycle, not 1 -> 4 -> 2 -> 1. As if you start with 7 and on some step get 7 again! Why why why it never actually happens?!!

Simon Multiverse Yıl önce

The cover of the book is taking a short cut. It has an arrow going from 1 to 2 so it's showing two steps n-> (3n+1)/2 (combining the multiplication and the division by 2).

maki 2 yıl önce

today was my first day at uni and my professor mentioned this conjecture. i was intrigued so i decided to look into it more THANKS NUMBERPHILE FOR COVERING IT

YASH RAWAT Yıl önce

*" Don't Judge a b̶o̶o̶k̶ problem by its c̶o̶v̶e̶r̶ statement "* *- Collatz Conjecture*

Robin Aylott Aylar önce

Genius!

SomeoneCommenting 5 yıl önce

If you make the inverse tree of the values starting from 1 up and checking every even integer to see if it can come from a previous even number divided by 2, or an odd number that was multiplied by 3 and had 1 added, you clearly see that _all_ integers will always follow a down path in the tree. No matter how high you can place a number in the tree. So why do mathematicians want a proof as a formula when the tree shows that the answer is YES so clearly? All numbers end up in 1.

Vexatos 6 yıl önce

And as we learned, Brady was most likely to choose 7 out of all numbers from 1 to 10.

some random bird 2 yıl önce

Nice, exactly 777 likes

danlock 3 yıl önce

@Hiromant Great idea! However, this question is stated two ways in this thread. "from 1 to 10" and "between 1 and 10"... which are not the same. You can answer 1 to the first but not to the second.

Hiromant 4 yıl önce

I always say 1 just to throw people off their game. Funny how nobody who asks it knows what random really is.

Rosyid Haryadi 4 yıl önce

maybe it was his birthday, or his lucky number.

Tim Russell 4 yıl önce

This function is chaotic but it does have a downward trend. So what is the average downward slope?

D O'Mota 4 yıl önce

it would be interesting to analyze the lenght of the branches

Guepardo Guepárdez 3 yıl önce

What would we get if we extended the funtion to complex numbers by considering them odd if the sum of their real and imaginary coefficients is odd and even otherwise? If we then filled every "pixel" (centered at a point with whole coefficients) in the plane with a different color depending on how long it takes to get 1 from its value by iterating the function, what kind of shape would we get? Would it be different from a coarse-grained picture of the Collatz fractal?

Yurii Sheliazhenko Yıl önce

In the article “Autonomous Version of Collatz Conjecture: Freedom of Choice Makes Unsolved Problem Solvable,” DOI: 10.20935/AL1999, the next theorem is proved. If an equivalence relation ↔ for any x∈N satisfies the rules 2x↔x and x↔3x+1, then x↔1 for any natural number x.

Griffon Theorist697 3 aylar önce

Thank you to TRvidr Engineering Made Easy for bringing this to my attention According to PR Newswire, the company Bakuage Co., a Japanese audio company, is offering a prize of 120 million Yen, which is worth a little more than a million US dollars. The article was released on July 7, 2021. I did read a joke on Reddit that the company may not be able to afford the prize, but beyond that I couldn't find anything legitimate debunking the claim.

Timothy Swan 6 yıl önce

You did it!! The Collatz! Thank you! Now, I am about to be away from electronics for 9 months. At least I'll know that you guys made a video on the conjecture like I asked. I tried to prove it as an exercise. Going to watch the extra footage, of course. :)

T. Kreischwurst Yıl önce

When there's a still not understood randomness underlying the Collatz conjecture, couldn't you use that conjecture for encryption purposes / for creating random numbers?

Cezar Catalin 11 aylar önce

It’s not exactly the most efficient way of screwing around with cryptanalysts.

Christoph Nimptsch Yıl önce

it is interesting to look at the problem with binary numbers. dividing by 2 is just binary shifting to the right, because lowest bit is 0 for even number. 3n+1 is shifting the initial number to the left, adding the original number and then add 1. the middle path is reached when the highest bit is 1 and the rest is 0. they then can be reduced to "1" by shifting to the right.

numberboxgamer 4 yıl önce

This is literally just "4 is cosmic" in math form lol I appreciate that very much.

Michael H 2 aylar önce

Gotta love those “very even” numbers! I hate the slightly even ones. His enthusiasm is great to listen to. Surely the parameters of the conjecture dictate this very behaviour? 3n+1 forces positive integers and n/2 allows mathematical gravity (integer flow descending to 1) to do the rest. It’s not hard to understand, but I’m not trying to show a proof and greater minds have struggled so it certainly is intriguing.

David Hahnert 9 aylar önce

I didn't know that mathematics had a Bob Ross. Homeboy mentions trees, clouds, and visual patterns in his explanation.

Ωμέγα(π) 4 yıl önce

I've seen some comments here, proposing 0 as a solution to the problem. Though, 0 is a whole number, the definition of Collatz Conjecture calls for positive integers, thus we can't use 0 as an example of a number that doesn't fall to 1, because 0 is neither positive or negative integer.

pig beners 2 yıl önce

Does it have to be 3n+1? Have variations been tried for Mn+1? It seems to be some sort of exploit in our number system, I wonder if equivalents could be drawn up and examined in binary, hex or octal

MuffinsAPlenty 2 yıl önce

Yes, other variants have been tried. 5n+1 has loops that don't include 1. Mn+1 for M an odd number at least 7 always has sequences which tend to infinity. You could ask the question, though, for an+b where a and b are odd numbers (and you can even have b be negative provided that it is small enough in magnitude). It has been proven that this general statement is undecidable. This means that there _cannot_ exist an algorithm which takes a and b as inputs and determines in a finite amount of time whether you always get to 1 for any n.

zkummer2359 4 yıl önce

Is there some actual value to 3n+1? I mean does it actually relate to something in reality like e or pi? Sort of seems like a parlor trick; one rule is designed to get you an even number, the other to get you odd numbers. Any time you get an even number that is an exponent of 2, the pattern collapses to 1. Not really much different than firing atoms randomly at a grid with numbered holes and then trying to figure out the pattern in all the numbers hit before an atom hit square #1. How can you expect to see a pattern in the results if you don't maintain a pattern in the math used to get those results?

Jess 5 yıl önce

I love watching your videos! They always give me something new to think about, as a future mathematician it's great to get a glimpse of what I could be working on!

Mat Rix 10 aylar önce

For any positive integer which can be reduced by the Collatz conjecture to 1, it can be proposed the special kind of equation. On the other hand If this equation can be created for the particular positive integer, this integer can be reduced by the Collatz conjecture to 1. (v1xr4 2105.0003) As another step it can be proven that such equation can be created for any particular positive integer.

Bram Weinreder 11 aylar önce

You have to cascade down from a number in 2^x, where x is an even number. Furthermore, the numbers that get there after nX3+1 have a very strong relation with prime numbers. It's no coincidence you ended on a prime before going back to the tree.

Aurelia Yıl önce

I mean... There are definitely some patterns like the "rail" on that cover of the book with all of the 2^x numbers, which of course goes straight to 1. So all multiples of two will do this, the only tricky parts are figuring out what multiples of 3, 5 and 7 will do.

Fish767 2 yıl önce

For anyone that does want to try this on your computer, set it up so it only tries odd numbers because an even number will become a smaller odd number and if the number resulting from doing these rules becomes less than your starting number you will have already tried it and it will fail. Also if the number is divisible by 4 right after an initial (3n+1) it fails because that would be (3n+1)/4 and that is always smaller than your starting number. I am going to try to solve this later. I have a fun idea around seeing how many times you can multiply by 3 and add 1 before it becomes divisible by 4. I think primes will be the best to work with.

Kyofu 4 yıl önce

The steps do have a pattern. There are never 2 odds in a row but the even numbers can have 1 or multiple in a row meaning for every 1 step forward you take 1 or more steps back, therefore you will always end at 1 regardless of the number you start at.

LAS MATEMATICAS MAS NUEVAS 2 yıl önce

Professor David Eisenbud estoy seguro de tener la demostración de dicha conjetura. Es resolvible y se cumple. En verdad hasta ahora ha existido mucha especulación pero pocas definiciones que expliquen su comportamiento en forma aceptada. De aqui a una semana la demostraré en mi canal

Geraldo Gomes 2 yıl önce

To prove Collatz conjecture we need to prove only three conditions: First the relationship has only one cycle and it is on "1"; second it don't diverge (so it converge) and third it converge to "1" . See my videos to see it's solution.

dragoda 4 aylar önce

Veritasium also made a great video about this conjuncture. I love math and your channel. Please never change.

Charles Johnson 2 yıl önce

I decided to try a graph approach to this, specifically digraphs, every vertex has two incoming edges, one representing 3n+1 for some n and one representing n/2 for some n. Additionally each vertex has one outgoing edge which goes to a vertex representing the collatz function being applied. Basically every number has two incoming edges and one outgoing edge

LOLFlyingPotatoes 4 yıl önce

The sequence of 2^n in the middle column is also connected to a 5, which is connected to a 3, and is then attached to a 7. This shows that all integers of the form 5(2^n), 3(2^n), 7(2^n) and 2^n are all going to collapse to one. If we can prove that at least all odd numbers end up in any of these numbers, we can prove that Collatz conjecture. The first thing is that every second number in the 5(2^n) pattern is of the form 3n + 1 where n is a positive integer. 10 reduces to 3, 40 reduces to 13, 160 reduces to 53, 640 reduces to 213. 2560 reduces to 853. They are all of the form (10(2^(2n)-1))/3. In the 7's, every second number is also of the form 3n + 1. 7 reduces to 2, 28 reduces to 9, 112 reduces to 37, 448 reduces to 149. These are all of the form (14(2^(2n)-1))/3. This shows that in x(2^n) where x is of either of the forms above, it will reduce down to one. This doesn't prove anything; feel free to mess around with it though.

Max Wazowski Yıl önce

It seems that this works not only with 3x + 1, but also with a polynomial of the form mx + 1, where m is an odd number, m∈N ... Let's take x = 1 and substitute odd numbers for the coefficient m: For m = 1, everything is reduced to the cycle 2 → 1 *; For m = 3, everything is clear; For m = 5 we come to 4 → 2 → 1; For m = 7 we also get 4 → 2 → 1 * But with m = 1 and x = 2 it will turn out 4 → 2 → 1 To test the values of m> 7 and x> 1, it would be nice to write a computer program ...

Gnc GenZ 5 yıl önce

The way I see the whole tree is that there are infinite vertical branches that start coming together but each vertical line is a number multiplied by 2 to the power of another number. If anyone can prove that all numbers can be written like this that might be the answer

TomKaren94 3 yıl önce

The 4-2-1 sequence has to be there for any number going to 1. Given the applied formulae rules, the only way to get to 1 is through 2, and the only way to get to 2 is through 4. The only way to get to 4 the FIRST time is through 8... etc. That's why the line defined by the powers of 2 are the "trunk" of the tree.

Mat Rix Yıl önce

For any positive integer which can be reduced by the Collatz conjecture to 1, it can be proposed the special kind of equation. On the other hand If this equation can be created for the particular positive integer, this integer can be reduced by the Collatz conjecture to 1. As another step it can be proven that such equation can be created for any particular positive integer. This all is proven in my work, which is available on vixra under the number 2105:0003. You can check on your own.

SuperDangerousMouse 8 aylar önce

apart from the hailstone phenomenon, are there other application of this Collatz Conjecture? Where else can we see it in nature?

Drenz 152 Yıl önce

Here is something I figured out: If the numbers will eventually end in a loop other than the 1-4-2 loop, it cant be in the pattern odd-even-odd-even... and so on, ending with an even and go back to the same odd number at the start. This might help!

BadHombre Yıl önce

This must be the formula that banks use, to calculate fees, that reduce my account to 1.

Mr. Ditkovich Yıl önce

🤣

Mitodru Misra Yıl önce

🤣🤣🤣🤣

J. F.G. Aylar önce

I think I may have found something interesting about the Collatz Conjecture, I don't know if it is enough to prove it but it's worth watching.

I Eat Garbage Yıl önce

I messed around with the problem a bit, and I found the lowest possible answer has to take form 16n+7, 16n+9, or 16n+11. Otherwise they would lead to a lower number that would have to also be a solution. I just got this by plugging in the forms that work, starting with 2n+1 (cuz the answer obvi isn't even), and finding what n must be in order for the equation to not spit out a lower number (in this case, n would have to take form 2n'+1), and plugging that in (in this case getting 4n'+3). I just quit when I saw the paths branch I'm also pretty sure you can just toss out any form you come across after plugging something in. When you plug in 2n+1, you get 6n+4. Any number in this form corresponds to a number in form 2n+1 that leads to it, meaning it would be a smaller solution. This would mean forms 3n+2, 6n+5, and 9n+8 are also off the table.

ezpotd 2 yıl önce

Also, has anyone proved it for x/2 for even and 3x-1 for odd

KingMRool 15 gün önce

I love this channel. This gives me some space vibes!

Q Squared 3 yıl önce

Essentially: N will always loop 1 to 4 for any value N, because for any given odd number O 3O+1 will be an even number, to which there will never be 0 divisions by 2, and so rarely be only 1 division by 2, that it must always trend downward. For any given Number N you have a finite number of paths which can lead to an odd number or an even number, going upwards you have an infinite number of paths which can lead to an even number, but making an even number that is not divisible by 2 is impossible, and making one that only can be divided by 2 only once is so much less likely to to occur than being divisible 2 or more times that that only 2 and 6 reach the state of always having only 1 further division, and 4 and 8 only in special cases, while all odd numbees beget a new even number, and the equation is open ended, so it will always trend to the 1 to 4 loop, how could it possibly not do so?

Fight Me 9 aylar önce

Id be curious to know the probability of an odd number input into 3x+1 and getting an even result, if that number is more than half the probability of a number being odd in general (50%) then it would prove a downwards trend wouldn't it?

Gregg Senne 3 yıl önce

I've found an interesting way to map the Collatz conjecture. Some interesting patterns emerge. I don't know what they mean, but they look kind of cool. I can share this with interested parties via email or Facebook.

Aaron Hollander 4 yıl önce

Rule #1 has a lowering effect on any number and outputs an even number a certain percentage of the time. While rule #2 increases the number, it always produces an even number which is reduced further. Seems obvious that the number would tend downward. One step forward and two steps back. The misdirection is that you're tripling the number half the time, that's simply not the case.

Legend 2 aylar önce

But if there was a number where the steps you take are like this: (3n+1)/2, (3n'+1)/2, (3n''+1)/2,... keep going like that forever, then the number would never reach 1.

Rajeev Bagra 5 aylar önce

Agree completely.

João Vitor Matos 5 yıl önce

This conjecture was in my maths test, it asked how many non-repeating primes was in the Collatz Conjecture. People failed because they thought 1 was prime

Michaele Pugliese 4 aylar önce

It is an absolute fact that all positive integers resolve to a multiple of 5, or a power of 2. I can easily show this beyond any doubt. As often as not it can take a many iterations to reach a multiple of 10 that is divisible by 2 many times, sometimes immediately iterating to a (2^n)*5 number, and sometimes reaching another cycle, through various ending digits, to get another multiple of 5 ending in a zero, iterating down again by dividing by 2. The beauty of this is that you cannot avoid ending up with a multiple of 10 which is divisible by 2 all the way to a number that is represented by (2^n)*5, which will always iterate to a power of 2. At any given moment a power of 2 itself can magically appear as well. It seems airtight to me, and I would welcome someone finding a flaw in my method. I do not have the advanced mathematical skills to submit a formal proof of my findings, but the patterns are so obvious that someone with those skills should be able to do some amazing things with my findings. I would love to collaborate with such an individual.

Griffon Theorist697 3 aylar önce

I do recognize 5 as an important root number in 3x+1, but I don't know if understanding this is enough to prove the Collatz Conjecture to be true. The first even number to take an even number steps to be odd will always be a power of 2 (4 takes 2 steps, 16 takes 4 steps, 64 takes 6 steps, etc.) and the first even number to take an odd number of steps to be odd will always be a power of 2 * 5 (10 takes 1 step, 40 takes 3 steps, 160 takes 5 steps, etc.). However, this is also true for the modified rule 3x+7. Even though 5 and 1 are still essential roots, this all gets thrown in the 5-11 loop. This logic by itself does not explain why the 28-14-7 loop exists. However, I think you make an interesting point. 3x+7 seems unusually stable. The Collatz clone loop is required and unavoidable, so it's kind of cheating to point to that as the "exception". I don't believe there are any other loops aside from the 5-11 loop. Apparently this loop must be huge because it swallows a bunch of negative numbers too. However, looking over 3x+13, my initial ideas on roots fall apart. Starting with 16 is a little weird, then 3 *3+13 gives you 22, so the first root is 11? But 5*3+13 gives you 28, the first divide by 2 number, so 7 is a root? 3x+13 is also insane, there are a lot of loops and whatever the heck is going on with its septuplet loops.

fyfferguy 2 yıl önce

To the OP: What is the source of the graph I see in the "collage" at 2:45? It's the one in the bottom center of the screen with the red dots on white background. This is exactly the data set that I have been playing with since I first heard of this problem in my college days over 25 years ago. I'd never seen the data presented that way before, and then I saw "my" graph in your video! Thanks in advance! (and hopefully someone will see this ... :( )

Russell Gokemeijer Yıl önce

I have seen that graph on the Wikipedia page for collate conjecture and I am sure they list a source

BucketCapacity 6 yıl önce

I have studied the Collatz Conjecture in my spare time. The closest thing I got to anything useful was, when n =/= 24*2^x+1, x in the integers, n going to 1 in the collatz function is equivalent to there existing integers a and b such that 4a | n+b and b | n +4a. This does work for certain n in the form of 24*2^x+1, but not all of them (ex n = 193)

PleaseDontWatchThese 5 yıl önce

Things always gets tricky when dividing odd numbers

Lkxemmeji 3 yıl önce

I came up with simple formula(I will use lower case for index, n is the number, p stands for confirmed amount of steps, Ap is the result of doing 3n+1 then divided by 2 exactly p times) Ap = 3^p * (n+1)/(2^p) - 1 which easily shows how to get as many steps as you want to (just get n+1 be properly divided by 2^p) and the amount of steps will be more than p. Though it doesn't solve anything :D

Hibbi 3 yıl önce

/2 is infinity - if you make a infinite x2 sequence - 1x2 - 2x2 - 4x2 - 8x2 - 16x2 - 32x2 - it happens to be the same numbers that drop the numbers straight back to 1 - so given that this infinite sequence of numbers exist, its infinitely certain it will eventually drop to 1, cause like infinity is a lot of numbers, like all of them

UK Rustafarian 3 yıl önce

Could it be said that 1 is just the attractor to a simple system?Half of all numbers are even and decrease by half. Half of those are even etc Is this not another way to frame 1n + 1/2n + 1/3n + 1/4n + 1/5n etc where n is probability of an even result? It could be rewritten with each fraction of probablity minusing the same amount from “m” probabiltity of an odd number. No matter where any number falls in the equation, if it is a continual then will allways add probability of even and reduce that of odd. So it is actually a strange attractor in a chaotic system because it is all drawn to 2 but the formation of equation makes it cycle 1, 4,2,1 etc Plus the powers of 2 all drop straight to 2 Like the riemann zeta function resting on 1/2.

DoodleFox 6 yıl önce

I like how it started like the fibonacci sequence; 1, 1, 2, 3, 5. I wonder if it follows that pattern even longer with a bigger tree..

SNBeast 4 yıl önce

Here's a tip if you're gonna write a program that looks sequentially for one that doesn't reduce: if you want to find the first example one that doesn't reduce, don't test even numbers. If you're looking for the first number, it's gonna be odd because an even number will divide to a smaller number. Since that number would have to not reduce, the number you're testing cannot be the first.

Channel Dad Bryon Lape 3 yıl önce

I wrote a version in C on an old Core 2 Duo I have running Linux and tried the 2^60 number in the video. Takes less than a second to spit out all the steps down to 1. I also tried numbers larger and they work very quickly too. Kinda surprised such a small number was given in the video.

Tristan Ridley Aylar önce

I know this is a 3 year old comment but... All powers of two will be incredibly fast, as they never grow. For your computer to solve that 2^x it only takes x iterations. Your computer literally only looped 60 times.

Rob Gross 11 aylar önce

But they've done EVERY number up to 2^(60) (further now). Suppose it took 0.001 s per number on average. That's 2^(57) s = 40 trillion hours = 1.7 trillion days = 4.5 billion years. Now, suppose they had 100,000 computers working on it at once that were 200 times faster than your computer. That's still 228 years to go through all the numbers.

Bill Streifer 5 yıl önce

I wish I had come up with this conjecture. The problem is, if I had confronted my math teacher with it, he probably would have asked me to come up with a counter-example for "extra credit." Oops!

Михаил Чувашов 3 yıl önce

There are 2 actions. In one case is divided into 2, and in another it is multiplied by 3 (+1 plays a role only of transfer of number to even-numbered after multiplication by 3 here). At alternation of multiplication and division of number, multiplication happens only once, but on 3, and division can repeat several times, though on two. It is necessary to resolve an issue (!) that in a limit average N of divisions into two will exceed one multiplication by 3, i.e. inequality has to be carried out N > log2(3), i.e. the average of repeated divisions on 2 has to be more ~ 1,585... And the average number of repeated divisions on 2 at all infinite set of numbers aspires to 2 in a limit that is obvious more, than 1,585. Therefore any number will be undoubted will decrease because division on 2 in average will be twice more, than multiplication by 3. As for multiplication by numbers since 5, the result will go to infinity, but not always. At those figures-multipliers where since 1 the recurrence will appear, there will be a certain set of numbers, the result of operations with which will lead to 1. There where the recurrence is absent, to 1 will give numbers 2^n, as well as in all other options. Excuse me for my English.

Martin Verrisin 4 yıl önce

I see why it tends towards 1: 2>1.5+c (for all big numbers and all small are checked) but I wonder what keeps it from ever entering a loop...

Red X 2 aylar önce

The numbers in the paths are on separate moduli. Take the number 7, and think of it as 128x+7 or 7 mod 128. Multiply by 3, add 1 and divide by 2 to get 192x+11 or 11 mod 192. Do that again, and you get 288x+17, then 216x+13, and 81x+5. Furthermore, you can cut out parts of these mods to show that the paths are part of even smaller moduli: 11 mod 192 is part of 11 mod 64. 17 mod 288 is part of 17 mod 32. And 13 mod 216 is part of 5 mod 8, leaving 5 mod 81 to be 2 mod 3 Because of this, we can clearly show the direction of the path along the branching tree of Collatz, and show that the number 7 (which we already knew) goes all the way to one. Generalizing it, we can show that any number above the point where we can verify has a matching pattern, a matching path, that lies within the numbers we can verify, and thus must also drop below itself to a number that has been verified, thus proving that all numbers, without exception, drop down to 1.

flurng Yıl önce

For those who didn't catch it already, the straight line that the professor refers to as the "3rd rail" is just successive powers of 2.

Qazsedc 2 aylar önce

AKA a *very* even number

wonggran 2 aylar önce

The collatz conjecture, only powers of two get to 1 and the conjecture's operators never get to 0. Therefore, only positive integers that are powers of 2 get to 1. So, there are positive integers that do not get to 1.

David Hatch 3 yıl önce

Every number to infinity will reduce to one in that equation set. They produce even numbers as their role. Matching every possible even number in existence will reduce to 1.

Seth Gilbertson 3 yıl önce

I use this pattern in my 4th grd math class! It’s awesome!

EFBensonFan 2 yıl önce

So if one proves that any integer you apply the algorithm to will eventually yield 2^2n (where n is equal to or larger than 1; e.g., 4, 16, 64, etc.) then one proves the conjecture. Easier said than done...

TheDaddyO44 5 yıl önce

I like this guy. Wish he'd been my maths teacher

Jeb_Kerm 3 yıl önce

What if you adapted this problem for complex numbers?

Spider Salticidae 2 yıl önce

For 3n - 1 , there are certain numbers that reach 1 and loop .

Kevin Gallagher 5 yıl önce

something that's always confused me is how exactly conjectures are "proven" or falsified. obviously, if you found one number that doesn't simply down to one then the conjecture is disproven, but, since numbers are infinite and we can't test them all, when can we possibly say that the conjecture has been proven to be true?

Bob Nob 5 yıl önce

there are ways. The easiest way to prove something for all natural numbers is mathematical induction. For example: n + 1 > n. We cant test all natural numbers but we could start with 1: 1 + 1 = 2 > 1 next we assume its correct for one natural number n and see if it holds true for n+1: (n + 1) + 1 > n + 1 n + 1 > n (which is our assumption) therefore we can conclude that if our statement is true for an arbitrary natural number n then its true for its successor. And since the natural numbers are arranged in a way that every number as a successor and we know its true for 1 we know that the statement is true for all natural numbers.

enlong chiou 4 yıl önce

Reciprocal of (3m+1)/2^n rule of Collatz conjecture is (2^n*m-1)/3, can use (2^n*m-1)/3=m for both, it's only solution is m=1.

H L 8 aylar önce

Please post more videos Enlong

Transyst 3 yıl önce

This only shows that it doesn't return to the same number after just one 3m+1 step, no matter how many consecutive /2 steps afterwards, except for m=1. But it doesn't exclude cycles with multiple 3m+1 steps, or the possibility that it doesn't end.

Misty Cremo 5 yıl önce

I feel that this problem may be easy enough to find a counter example for if you could search for loops without testing any number. Maybe try plugging in 3(2x)+1=x, or other equations that show the process that might occur to any number, and if you find that a number that would, following the rules of the conjecture, normally be able to follow such a pattern, you would find a loop that hopefully does not end in 1.

Alan Tyte Yıl önce

There is a bottom up structure for Collatz that includes all numbers to infinity but it would upset too many people for it to be taken seriously.

Fujtajblus 2 yıl önce

I tried something different. I took 0,11 and it keeps going up. I think the problem talks about using whole numbers, right? If you use any number between 0 and 1 and go trough this cycle will eventually go up.

Olixx12 6 aylar önce

How do you know if 0.11 is even or odd , makes no sense . Thats why its for strictly positive number

Michael Gmirkin 3 yıl önce

So, I wonder has anybody come at the primary "3n+1" problem from a slightly different angle of trying to prove for specific subsets of starting numbers, that they will "always go to 1"? For instance, has anyone ever tried to prove that, say, for just multiples of 3, "all multiples of 3, will always end up at 1, via the rule(s)"? (Given than multiples of 3 will NEVER have a "3n+1" input, but only "n/2" inputs, so they're kind of an interesting set.)? Or like, multiples of 5, etc.? Like some specific subsets that might give insight into the problem, even if they don't prove the conjecture over the set of **all** possible positive integers? I mean, we already know that any powers of 2 (e.g., 2^n) will always collapse to 1. Do we know if there are any other classes of numbers, or multiples of specific numbers that, for sure, will "always" collapse to 1? Even if it doesn't prove the conjecture as a whole? Just a thought I had... Just wondering... ^_^

Kevin Gil 2 yıl önce

0:16 He chose seven on purpose! He knows it's the least random number! Bold strategy.

Reziik 6 yıl önce

I just learned to make simple stuff in python yesterday and my second program was making a loop to take a random number between a set range and do this.

CASPERA 3 yıl önce

I do not really understand how it is possible to prove these kind of patterns mathematically. Have we ever solved a somewhat similar problem?

Q Squared 3 yıl önce

Okay, but, obviously, inherently, any even whole number, when divided by 2, will reach an odd number, if you always multiply that number by 3 and add a 1 to it you will end up wirh a new even number, which is larger than the odd number that spawned it but all even numbers not ending in a 2, or a 6, will always divide at least twice by 2, meaning your over-all trend MUST be towards 1, and the 1 to 4 loop. There can be no other outcome, only the edge case of 10 breaks the 2/6 rule, and that is already proven to drop to 1 anyway. I fail to see how this isn't provable? what about my reasoning is not writable logically? It is impossible to cause a sequence of 3n+1 that leads to a 2 or a 6 that would after being divided by 2 would lead to another 2 or 6 when multiplied by 3 and having 1 added!

Andrew Bounds 3 yıl önce

Now I'm pretty sure that all the old numbers (to the point where you must go to 3x+1) are all primes

RPG Llama 2 yıl önce

Can you generalize this to other numbers. By which I mean, can you make a version which boils down to powers of 3 or 5 or 6?

MuffinsAPlenty 2 yıl önce

In theory, you could. It would probably depend on "remainder" though. For example, if you want to deal with powers of 3, you could set up a rule like this: If n is a multiple of 3, divide by 3. If n is 1 more than a multiple of 3, do 5n+1 (which will then be a multiple of 3, since if n = 3k+1, then 5(3k+1)+1 = 15k+6 = 3(5k+2)). If n is 2 more than a multiple of 3, do 5n+2 (which will then be a multiple of 3, since if n = 3k+2, then 5(3k+2)+2 = 15k+12 = 3(5k+6)). I made some choices in this procedure. You could make other choices, though. For example, you could have a sort of cycle, where numbers 1 more than a multiple of 3 are sent to numbers 2 more than a multiple of 3, which are then sent to multiples of 3.

Daniel Röder 3 yıl önce

If there was one number that wouldn't go down to 1 it means there must be at least a bunch of them, because whatever this number is, following the rules you would go to other numbers. So the only possible way that there were numbers that don't go down to 1 is when there is a closed loop of numbers that are not part of the tree, so that when you start with one of them you go through the loop and get back to your starting number somehow. Edit: Only other way would be infinite numbers where one is the smallest and following it you may only get to bigger numbers, and never get to a number smaller than the first one.

Q Squared 3 yıl önce

Trying to form what I see in my head into words is hard. This maybe this is clearer. The set of even whole numbers which can be divided by 2 more than once is larger than ths set of even whole numbers which can be divided by 2 only once. OR to put it another way the set of numbers which are multiples of 4 is the SAME size as the set of numbers which are multiples of 2, but not multiples of 4! And therefore, you will get as many divisions twice (or more times) as you will get divisions once meaning that 2/3 of the time your pressure on the dervived number will swing you downward, and since 4 is the largest factor of merit, you will always reach 4 which will then begin your loop of 1 to 4. Note that this is explicitly due to the factor chosen being 3n+1. IE: 4 is a multiple of 4, 8, 12, 16, 20, 24, 28, 32 2 is a multiple of 2, 6, 10, 14, 18, 22, 26, 30, 34 as you can see these sets are exclusive, and they are both infinite, but they must be of equal sets. look at it this way. for any given equal set of odd and even numbers, the set of all odd numbers multiplied by 2 is equal to the set of all even numbers multiplied by 2. which means even for finite values this will be true, so any given even number will be able to be devided by 4 50% of the time, and by 2 50% of the time, and that is a probability, so you will always have a 50% chance that you will be able to devide an even number by 2 again once you device it the 1st time, always.

Neil Ruedlinger 3 yıl önce

I'm not sure if anyone noticed this, but in the tree from Wikipedia, I can only see two trifurcations, the rest are bifurcations. When I first noticed the bifurcations, I got a bit excited thinking that there may be a link with Fractals and Chaos Theory, since one characteristic of a Chaotic System is the tendency for a key parameter to bifurcate, then I noticed the two trifurcations and became a bit dismayed...

vidar traeland 2 yıl önce

Could the collatz conjecture be proved in the following way?: «Prove that there can NOT be a number >1 that neither eventually leads to, or is followed by a number lower than itself.» The logic is that if any numbers contradicts the conjecture, there has to be a number with this property. (The lowest number that does not lead to 1, can not have a lower number before or after)

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